Bounded integral of absolute value of derivative squared

Old qual question:

Let $f$ be a $C^1$ function on $[0,1]$ and assume $$\int_0^1 |f'(s)|^2\,ds\leq A.$$ Show that, for all $x,y\in[0,1]$, $$|f(x)-f(y)|\leq A^{1/2}|x-y|^{1/2}.$$

I really don't know where to go with this one. I'm guessing using the mean value theorem somewhere, but I don't know.

• FTC and Cauchy-Schwarz. – Daniel Fischer Dec 14 '16 at 14:42

The basic idea is using the fundamental theorem of calculus, we have $$f(x) - f(y) = \int_y^x f'(\xi)\, d\xi$$ The triangle inequality gives, wlog $y<x$ $$\def\abs#1{\left|#1\right|}\abs{f(x)-f(y)}\le \int_y^x \abs{f'(\xi)}\,d\xi$$ The Cauchy-Schwartz inequality applied for $\abs{f'}$ and $1$, gives $$\int_y^x \abs{f'(\xi)}\, d\xi\le \left(\int_y^x \abs{f'(\xi)}^2\, d\xi\right)^{1/2} \left(\int_y^x \abs{1}^2\,d\xi\right)^{1/2} \le A^{1/2}\abs{x-y}^{1/2}$$